# Tag Info

Accepted

### Multiplication by a Haar random unitary two times

There is an explicit formula for the integral with respect to the Haar measure of any polynomial in the entries of a unitary and its conjugate, due to Collins and Śniady: Benoît Collins and Piotr ...
Accepted

### Is there a lower bound on the average diamond norm of two uniformly random unitaries U1 and U1 of dimension D that are sampled from haar measure?

This answer won't actually give you a bound, but will provide some information that may help you in your search. You may be able to find an answer in the random matrix theory literature if you ...

### What is a Haar random quantum state?

Typically this is a slight abuse of notation. One can have a unitary operator $U$ chosen from some Haar measure, such as the circular unitary ensemble. Then, taking some fiducial state $|\psi_0\rangle$...

### What is the expectation value ${\Bbb E}[\langle\psi,O\psi\rangle]$ over the Haar distribution?

Since a Haar-random $\lvert\psi\rangle=U\lvert0\rangle$ for a Haar-random $U$, your expectation value equals $$\langle 0 \rvert \Big[\int \mathrm d U\, UOU^\dagger\Big]\lvert0\rangle\ .$$ The ...

### Is the Haar measure invariant under conjugation?

I will answer this question in a more general context. You might know that Haar's theorem tells you that on any locally compact group $G$, there is a unique left-invariant (Borel) measure $\mu$, up to ...

### Is the column vector of a uniformly sampled random unitary matrix a uniformly sampled random state vector?

Suppose that was not the case. Then taking the first column of a uniformly random unitary matrix gives you a nonuniformly random state. That means that there is some state, call it $|\psi\rangle$, ...

### Is the column vector of a uniformly sampled random unitary matrix a uniformly sampled random state vector?

Yes. A uniformly (Haar random) sampled state vector $|\psi\rangle$ is characterized by the fact that the probability measure is invariant under any $U$, i.e., colloquially, $U|\psi\rangle$ is just as ...
Accepted

Accepted

### Random quantum states and Schur-Weyl duality

Note that the quoted relation $$\bar M_i = \sum_\lambda a_\lambda P_\lambda,$$ only holds if the $M_i$ also commute with the representation of the symmetric group! Otherwise this can obviously not ...

### Spoofing XQUATH with the Feynman method

The paper does not specify the exact algorithm or class of distributions $\mathcal{D}$ for which such algorithm fails to refute XQUATH, and some classes of distributions $\mathcal{D}$ do not satisfy ...
Accepted

### Quantum hardness of XQUATH conjecture

Maybe think of it this way - a quantum computer, executing a small enough random circuit $C$ acting on a state initially prepared as $\vert 0^n\rangle$ and sampling therefrom, will get an $n$-bit ...
Accepted

### Quantum supremacy: shallow depth Haar random circuits and unitary designs

First of all, that does not imply anything for shorter (constant/logarithmic) depths. Moreover, the 2-design property does not imply that the outcome distribution is the same as for Haar-random ...

Accepted

### Average output state of random quantum circuits

Calculating $\rho_1$ Let $N=2^n$ denote the dimension of the Hilbert space where $|\psi\rangle$ lives. For $i=0,\dots,N-1$, let $V_i$ be any unitary that maps $|i\rangle$ to $|0\rangle$. The action of ...
1 vote

### Sampling Haar over two systems

I would choose to think of $|\psi\rangle$ as $U|0\rangle$ where $U$ is any unitary. But, I can also think of it as $U'|1\rangle$, or $U''|2\rangle,\ldots$. Hence, I can write this as \begin{align*} \...
1 vote

### Question regarding integration of Haar random state

I was also flummoxed by the apparent puzzle why the value of $\mathbb{E}[\langle z|C|0^n\rangle|^2]$ is $2/2^n$ instead of $1/2^n$, but in my opinion I think the confusion arises from the symbol \$\...

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